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Divergence of a Curl - Then Integrate By Parts!

  1. Oct 21, 2012 #1
    1. The problem statement, all variables and given/known data
    ∫Bdot[∇×A]dV=∫Adot[∇×B]dV

    Prove this by integration by parts. A(r) and B(r) vanish at infinity.


    2. Relevant equations
    I'm getting stuck while trying to integrate by parts - I end up with partial derivatives and dV, which is dxdydz?


    3. The attempt at a solution

    I can break things down to Cartesian components, but integrating by parts is where I get stuck.

    Essentially, I'm simplifying by stating the identity that
    BdotCurlA - AdotCurlB = Div(A×B) = 0 in this case (Subtract right hand side from left and combine under one ∫dV)

    The components look like this:
    ∂x(AyBz−AzBy)=(∂xAy)Bz+Ay(∂xBz)−(∂xAz)By−Az(∂xBy)
    plus the y,z terms as well.

    How would one integrate ∫[(∂xAy)Bz]dV by parts? I have Ay and Bz which are potentially functions of x,y,z and a partial wrt x, and dxdydz...

    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 21, 2012 #2

    dextercioby

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    Do you know cartesian tensors ? If so, the problem is trivial.

    [tex] \iiint A_k \epsilon_{klm} \partial_{l}B_{m} dV = ... [/tex]
     
  4. Oct 21, 2012 #3

    lurflurf

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    Integrations by parts are equivalent to product rules.
    use
    [tex]\nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B})[/tex]
    and the divergence theorem to reduces the problem to showing a surface integral at infinity is zero.
     
  5. Oct 22, 2012 #4
    No, we haven't studied those so I don't believe that's a path I can take.

    I have used that identity to reduce the problem to ∫Div(A×B)=0

    This reduces to:
    ∫[∂xAy*Bz-∂xAz*By+∂yAz*By-∂yAx*BZ+∂ZAx*By-∂zAy*Bx]dV
     
  6. Oct 22, 2012 #5
    So I guess I'm wondering how to handle the partial derivatives inside the integrand. Take the first term for example:
    ∫∂/∂x(AyBz)dxdydz

    Do the ∂x and dx cancel out?
     
  7. Oct 22, 2012 #6

    dextercioby

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    No, no, no. Do you know about the Gauß-Ostrogradskii theorem ? It turns a volume integral into a surface integral.
     
    Last edited: Oct 22, 2012
  8. Oct 22, 2012 #7
    Yes, but I was focused on the brute force of integration by parts method as opposed to using this theorem. I think I'm just going to go the easy route and use this method - hopefully I won't lose too many points!
     
  9. Oct 22, 2012 #8

    dextercioby

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    You are doing vector integrals, you should use the vectors not their components, and you can use the Gauß-Ostrogradskii theorem in case of vectors.

    <In vector calculus, the divergence theorem, also known as Gauss's theorem or Ostrogradsky's theorem,[1] is a result that relates the flow (that is, flux) of a vector field through a surface to the behavior of the vector field inside the surface.>

    No reason to lose points.
     
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